Fourier Transform Syntax and Conventions Clarification












0














One common definition of a fourier transform for function f(x) is
$$F(v)=int_{-infty}^infty f(tau)e^{2pi itau v}dtau$$
I know some definitions have an extra sqrt(2*pi). I shall ignore that.



However, there are 3 tau's in the formula and it does not specify how each one relates to x. For example it does not explain for the fourier transform of f(x+1), whether tau+1 replaces all the taus or only for some.



From looking at derivations of fourier transform properties I believe that you must state state both the function and the variable the fourier transform is of. However, if your function is say f((y-1)) convention assumes that the variable is y instead of say (y-1).




  1. Is that right?

  2. If so, is this the correct fourier transform for f(g(x)) with respect to x?
    $$F(v)=int_{-infty}^infty f(g(tau))e^{2pi i tau v}dtau$$










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  • take a look here en.wikipedia.org/wiki/Fourier_transform#Other_conventions
    – alexjo
    2 days ago










  • @alexjo Yes, I am aware of those conventions. However, all those conventions are defined for f(x). Not for f(x+1) or f(g(x))
    – roobee
    2 days ago
















0














One common definition of a fourier transform for function f(x) is
$$F(v)=int_{-infty}^infty f(tau)e^{2pi itau v}dtau$$
I know some definitions have an extra sqrt(2*pi). I shall ignore that.



However, there are 3 tau's in the formula and it does not specify how each one relates to x. For example it does not explain for the fourier transform of f(x+1), whether tau+1 replaces all the taus or only for some.



From looking at derivations of fourier transform properties I believe that you must state state both the function and the variable the fourier transform is of. However, if your function is say f((y-1)) convention assumes that the variable is y instead of say (y-1).




  1. Is that right?

  2. If so, is this the correct fourier transform for f(g(x)) with respect to x?
    $$F(v)=int_{-infty}^infty f(g(tau))e^{2pi i tau v}dtau$$










share|cite|improve this question







New contributor




roobee is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




















  • take a look here en.wikipedia.org/wiki/Fourier_transform#Other_conventions
    – alexjo
    2 days ago










  • @alexjo Yes, I am aware of those conventions. However, all those conventions are defined for f(x). Not for f(x+1) or f(g(x))
    – roobee
    2 days ago














0












0








0







One common definition of a fourier transform for function f(x) is
$$F(v)=int_{-infty}^infty f(tau)e^{2pi itau v}dtau$$
I know some definitions have an extra sqrt(2*pi). I shall ignore that.



However, there are 3 tau's in the formula and it does not specify how each one relates to x. For example it does not explain for the fourier transform of f(x+1), whether tau+1 replaces all the taus or only for some.



From looking at derivations of fourier transform properties I believe that you must state state both the function and the variable the fourier transform is of. However, if your function is say f((y-1)) convention assumes that the variable is y instead of say (y-1).




  1. Is that right?

  2. If so, is this the correct fourier transform for f(g(x)) with respect to x?
    $$F(v)=int_{-infty}^infty f(g(tau))e^{2pi i tau v}dtau$$










share|cite|improve this question







New contributor




roobee is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











One common definition of a fourier transform for function f(x) is
$$F(v)=int_{-infty}^infty f(tau)e^{2pi itau v}dtau$$
I know some definitions have an extra sqrt(2*pi). I shall ignore that.



However, there are 3 tau's in the formula and it does not specify how each one relates to x. For example it does not explain for the fourier transform of f(x+1), whether tau+1 replaces all the taus or only for some.



From looking at derivations of fourier transform properties I believe that you must state state both the function and the variable the fourier transform is of. However, if your function is say f((y-1)) convention assumes that the variable is y instead of say (y-1).




  1. Is that right?

  2. If so, is this the correct fourier transform for f(g(x)) with respect to x?
    $$F(v)=int_{-infty}^infty f(g(tau))e^{2pi i tau v}dtau$$







fourier-transform convention






share|cite|improve this question







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roobee is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question







New contributor




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Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question






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Check out our Code of Conduct.









asked 2 days ago









roobee

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New contributor




roobee is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor





roobee is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






roobee is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • take a look here en.wikipedia.org/wiki/Fourier_transform#Other_conventions
    – alexjo
    2 days ago










  • @alexjo Yes, I am aware of those conventions. However, all those conventions are defined for f(x). Not for f(x+1) or f(g(x))
    – roobee
    2 days ago


















  • take a look here en.wikipedia.org/wiki/Fourier_transform#Other_conventions
    – alexjo
    2 days ago










  • @alexjo Yes, I am aware of those conventions. However, all those conventions are defined for f(x). Not for f(x+1) or f(g(x))
    – roobee
    2 days ago
















take a look here en.wikipedia.org/wiki/Fourier_transform#Other_conventions
– alexjo
2 days ago




take a look here en.wikipedia.org/wiki/Fourier_transform#Other_conventions
– alexjo
2 days ago












@alexjo Yes, I am aware of those conventions. However, all those conventions are defined for f(x). Not for f(x+1) or f(g(x))
– roobee
2 days ago




@alexjo Yes, I am aware of those conventions. However, all those conventions are defined for f(x). Not for f(x+1) or f(g(x))
– roobee
2 days ago










1 Answer
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The Fourier transform of $f(x)$ is
$$
mathcal F{f(x)}=F(nu)=int_{-infty}^infty f(x),mathrm e^{-2pi i x nu},mathrm d x
$$

With this convention, the inverse Fourier transform is
$$
mathcal F^{-1}{F(nu)}=f(f)=int_{-infty}^infty F(nu),mathrm e^{+2pi i x nu},mathrm d nu
$$

For $f(x+1)$ we have (time shift property)
$$
mathcal F{f(x+1)}=int_{-infty}^infty f(x+1),mathrm e^{-2pi i (x+1) nu},mathrm d x=F(nu),mathrm e^{+2pi i nu}
$$

For $f(g(x))=h(x)$ we have to evaluate
$$
mathcal F{f(g(x))})=mathcal F{h(x)}=H(nu)=int_{-infty}^infty f(g(x)),mathrm e^{-2pi i xnu},mathrm d x
$$

There is no genral rule in this case. If $g$ is a bijection and smooth enough then, if integral exists
$$
mathcal F{f(g(x))}=int_{-infty}^infty f(g(x)),mathrm e^{-2pi i xnu},mathrm d x=int_{-infty}^infty f(y),mathrm e^{-2pi i g^{-1}(y)nu}frac{1}{|g'(y)|},mathrm d y
$$






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  • You answered question 2 but not did not directly answer question 1. However, judging by your answer I shall assume you answered yes to question 1.
    – roobee
    2 days ago











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









0














The Fourier transform of $f(x)$ is
$$
mathcal F{f(x)}=F(nu)=int_{-infty}^infty f(x),mathrm e^{-2pi i x nu},mathrm d x
$$

With this convention, the inverse Fourier transform is
$$
mathcal F^{-1}{F(nu)}=f(f)=int_{-infty}^infty F(nu),mathrm e^{+2pi i x nu},mathrm d nu
$$

For $f(x+1)$ we have (time shift property)
$$
mathcal F{f(x+1)}=int_{-infty}^infty f(x+1),mathrm e^{-2pi i (x+1) nu},mathrm d x=F(nu),mathrm e^{+2pi i nu}
$$

For $f(g(x))=h(x)$ we have to evaluate
$$
mathcal F{f(g(x))})=mathcal F{h(x)}=H(nu)=int_{-infty}^infty f(g(x)),mathrm e^{-2pi i xnu},mathrm d x
$$

There is no genral rule in this case. If $g$ is a bijection and smooth enough then, if integral exists
$$
mathcal F{f(g(x))}=int_{-infty}^infty f(g(x)),mathrm e^{-2pi i xnu},mathrm d x=int_{-infty}^infty f(y),mathrm e^{-2pi i g^{-1}(y)nu}frac{1}{|g'(y)|},mathrm d y
$$






share|cite|improve this answer





















  • You answered question 2 but not did not directly answer question 1. However, judging by your answer I shall assume you answered yes to question 1.
    – roobee
    2 days ago
















0














The Fourier transform of $f(x)$ is
$$
mathcal F{f(x)}=F(nu)=int_{-infty}^infty f(x),mathrm e^{-2pi i x nu},mathrm d x
$$

With this convention, the inverse Fourier transform is
$$
mathcal F^{-1}{F(nu)}=f(f)=int_{-infty}^infty F(nu),mathrm e^{+2pi i x nu},mathrm d nu
$$

For $f(x+1)$ we have (time shift property)
$$
mathcal F{f(x+1)}=int_{-infty}^infty f(x+1),mathrm e^{-2pi i (x+1) nu},mathrm d x=F(nu),mathrm e^{+2pi i nu}
$$

For $f(g(x))=h(x)$ we have to evaluate
$$
mathcal F{f(g(x))})=mathcal F{h(x)}=H(nu)=int_{-infty}^infty f(g(x)),mathrm e^{-2pi i xnu},mathrm d x
$$

There is no genral rule in this case. If $g$ is a bijection and smooth enough then, if integral exists
$$
mathcal F{f(g(x))}=int_{-infty}^infty f(g(x)),mathrm e^{-2pi i xnu},mathrm d x=int_{-infty}^infty f(y),mathrm e^{-2pi i g^{-1}(y)nu}frac{1}{|g'(y)|},mathrm d y
$$






share|cite|improve this answer





















  • You answered question 2 but not did not directly answer question 1. However, judging by your answer I shall assume you answered yes to question 1.
    – roobee
    2 days ago














0












0








0






The Fourier transform of $f(x)$ is
$$
mathcal F{f(x)}=F(nu)=int_{-infty}^infty f(x),mathrm e^{-2pi i x nu},mathrm d x
$$

With this convention, the inverse Fourier transform is
$$
mathcal F^{-1}{F(nu)}=f(f)=int_{-infty}^infty F(nu),mathrm e^{+2pi i x nu},mathrm d nu
$$

For $f(x+1)$ we have (time shift property)
$$
mathcal F{f(x+1)}=int_{-infty}^infty f(x+1),mathrm e^{-2pi i (x+1) nu},mathrm d x=F(nu),mathrm e^{+2pi i nu}
$$

For $f(g(x))=h(x)$ we have to evaluate
$$
mathcal F{f(g(x))})=mathcal F{h(x)}=H(nu)=int_{-infty}^infty f(g(x)),mathrm e^{-2pi i xnu},mathrm d x
$$

There is no genral rule in this case. If $g$ is a bijection and smooth enough then, if integral exists
$$
mathcal F{f(g(x))}=int_{-infty}^infty f(g(x)),mathrm e^{-2pi i xnu},mathrm d x=int_{-infty}^infty f(y),mathrm e^{-2pi i g^{-1}(y)nu}frac{1}{|g'(y)|},mathrm d y
$$






share|cite|improve this answer












The Fourier transform of $f(x)$ is
$$
mathcal F{f(x)}=F(nu)=int_{-infty}^infty f(x),mathrm e^{-2pi i x nu},mathrm d x
$$

With this convention, the inverse Fourier transform is
$$
mathcal F^{-1}{F(nu)}=f(f)=int_{-infty}^infty F(nu),mathrm e^{+2pi i x nu},mathrm d nu
$$

For $f(x+1)$ we have (time shift property)
$$
mathcal F{f(x+1)}=int_{-infty}^infty f(x+1),mathrm e^{-2pi i (x+1) nu},mathrm d x=F(nu),mathrm e^{+2pi i nu}
$$

For $f(g(x))=h(x)$ we have to evaluate
$$
mathcal F{f(g(x))})=mathcal F{h(x)}=H(nu)=int_{-infty}^infty f(g(x)),mathrm e^{-2pi i xnu},mathrm d x
$$

There is no genral rule in this case. If $g$ is a bijection and smooth enough then, if integral exists
$$
mathcal F{f(g(x))}=int_{-infty}^infty f(g(x)),mathrm e^{-2pi i xnu},mathrm d x=int_{-infty}^infty f(y),mathrm e^{-2pi i g^{-1}(y)nu}frac{1}{|g'(y)|},mathrm d y
$$







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share|cite|improve this answer



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answered 2 days ago









alexjo

12.3k1329




12.3k1329












  • You answered question 2 but not did not directly answer question 1. However, judging by your answer I shall assume you answered yes to question 1.
    – roobee
    2 days ago


















  • You answered question 2 but not did not directly answer question 1. However, judging by your answer I shall assume you answered yes to question 1.
    – roobee
    2 days ago
















You answered question 2 but not did not directly answer question 1. However, judging by your answer I shall assume you answered yes to question 1.
– roobee
2 days ago




You answered question 2 but not did not directly answer question 1. However, judging by your answer I shall assume you answered yes to question 1.
– roobee
2 days ago










roobee is a new contributor. Be nice, and check out our Code of Conduct.










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